php通过引用传递不工作

Question

I'm just trying to understand passing by reference in PHP by trying some examples found on php.net. I have one example right here found on the php website but it does not work:

function foo(&$var)
{
    return $var++;
}

$a=5;
echo foo($a); // Am I not supposed to get 6 here? but I still get 5

This is an example found here

Can anybody tell me why am I getting 5 instead of 6 for the variable $a?

Answer 1

Your code and the example code is not the same. Is it a wonder they behave differently?

To see the behavior you expect, you have to change $var++ to ++$var.

What happens here is that while the value of $a is 6 after the function returns, the value that is returned is 5 because of how the post-increment operator ($var++) works. You can test this with:

$a=5; 
echo foo($a); // returns 5
echo $a; // but prints 6!